A very interesting question I am trying to solve in Thurstons 3 dimensional geometry and topology I found a pretty interesting problem in a wonderful book on three dimensional topology by Thurston. I have been thinking about the problem for awhile now, and am a little stumped. Here are the two questions to this problem:


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*Show that the linkage in the picture performs as advertised in its caption.

*Construct a mechanical linkage that achieves straight line motion. 
What do you guys think? I have plenty of rough sketches for the second one however, I am still a little unsure. As for the first, I was not sure how to go about this problem. 

 A: Note that the text underneath the diagram gives you the radius of the circle in which the inversion happens. I've hidden the computations in case you want to have a go yourself. The mechanism for drawing a straight line is known as the Peaucellier Linkage.
Let $O$ be the anchor point marked by the triangle, $C$ be the centre of the diamond and $A$ and $B$ be the unmarked vertices of the diamond.
The condition for inversion is then $OP\cdot OP'=r^2=x^2-y^2$. We can verify this from the diagram:

$$x^2=OA^2=OC^2+CA^2=(OP+PC)^2+CA^2=OP^2+2OP\cdot PC+(PC^2+CA^2)$$$$=OP\cdot(OP+2PC)+y^2=OP\cdot OP'+y^2$$

For part $2$

A circle through $O$ inverts into a straight line - so we need to find a way of constraining $P$ to part of a circle through $O$. Then $P'$ will trace a straight line segment. Select point $D$ as a second anchor point, and join $P$ to $D$ using a straight rod of length $l=|OD|$ so that $P$ lies on the circle centre $D$ and radius $l$, which by construction passes through $O$.

